Question: Let $y=\dfrac{x^2+2x}{4-5x}$. What is the value of $\dfrac{dy}{dx}$ at $x=2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac19$ (Choice B) B $-\dfrac23$ (Choice C) C $-\dfrac43$ (Choice D) D $-\dfrac65$
Solution: Let's first find the expression for $\dfrac{dy}{dx}$ (i.e. for any input value $x$ ). Then, we can plug $x=2$ and evaluate. $\dfrac{x^2+2x}{4-5x}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d y d x = d d x ( x 2 + 2 x 4 − 5 x ) = ( 4 − 5 x ) d d x ( x 2 + 2 x ) − ( x 2 + 2 x ) d d x ( 4 − 5 x ) ( 4 − 5 x ) 2 The quotient rule = ( 4 − 5 x ) ( 2 x + 2 ) − ( x 2 + 2 x ) ( − 5 ) ( 4 − 5 x ) 2 Differentiate ( x 2 + 2 x ) & ( 4 − 5 x ) = 8 x + 8 − 10 x 2 − 10 x + 5 x 2 + 10 x ( 4 − 5 x ) 2 Expand = − 5 x 2 + 8 x + 8 ( 4 − 5 x ) 2 \begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{x^2+2x}{4-5x}\right) \\\\ &=\dfrac{(4-5x)\dfrac{d}{dx}(x^2+2x)-(x^2+2x)\dfrac{d}{dx}(4-5x)}{(4-5x)^2} \gray{\text{The quotient rule}} \\\\ &=\dfrac{(4-5x)(2x+2)-(x^2+2x)(-5)}{(4-5x)^2} \gray{\text{Differentiate }(x^2+2x)\text{ & }(4-5x)} \\\\ &=\dfrac{8x+8-10x^2-10x+5x^2+10x}{(4-5x)^2} \gray{\text{Expand}} \\\\ &=\dfrac{-5x^2+8x+8}{(4-5x)^2} \end{aligned} So we found that $\dfrac{dy}{dx}=\dfrac{-5x^2+8x+8}{(4-5x)^2}$. Now let's plug $x= 2$ : $\begin{aligned} &\phantom{=}\dfrac{-5( 2)^2+8( 2)+8}{(4-5( 2))^2} \\\\ &=\dfrac{-20+16+8}{(-6)^2} \\\\ &=\dfrac19 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $x=2$ is $\dfrac19$.